Question

10.    If the roots ff the equation

           (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that

           either a = 0 or a3 + b3 + c3 = 3abc

Answer 1

hello friend here is your answer hope it helps u

Answer 2

Given Equation is (c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0.

Here, we have a = (c^2 - ab), b = -2(a^2 - bc) , c = b^2 - ac.

Given that the equation has real roots.

⇒ D = 0

⇒ b^2 - 4ac = 0

Putting values, we get

⇒ [-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0

⇒ 4(a^4 + b^2c^2 - 2a^2bc) - 4(b^2c^2 - ac^3 - ab^3 + a^2bc) = 0

⇒ 4(a^4 + b^2c^2 - 2a^2bc) - 4b^2c^2 + 4ac^3 + 4ab^3 - 4a^2bc = 0

⇒ 4a^4 + 4b^2c^2 - 8a^2bc - 4b^2c^2 + 4ac^3 + 4ab^3 - 4a^2bc = 0

⇒ 4a^4 + 4ab^3 + 4ac^3 - 12a^2bc = 0

⇒ a^4 + ab^3 + ac^3 - 3a^2bc = 0

⇒ a(a^3 + b^3 + c^3 - 3abc) = 0

⇒ a = 0, (or) a^3 + b^3 + c^3 - 3abc = 0

⇒ a = 0 (or) a^3 + b^3 + c^3 = 3abc.

Hope it helps!